4 Marks Questions

Question 14 Marks

A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL^-1, then what shall be the molarity of the solution?

Answer

10% (w/w) glucose solution means that 10 g of glucose is present in 100 g of solution which contains 90 g of water,
(i) Hence the molality of the solution (m)
$=\frac{\text { Weight of solute }( g )}{\text { Molar mass of solute } \times \text { Mass of solvent }( kg )}$
$=\frac{10}{180 \times 90 \times 10^{-3}}$
$\left[\begin{array}{l}\text { Molar mass of glucose }\left( C _6 H _{12} O _6\right)=180 \\ \text { Mass of solvent }=90 g=90 \times 10^{-3} kg\end{array}\right]$
= 0.617 m
(ii) Moles of glucose $=\frac{10}{180}$
Mole of water $=\frac{90}{18}=5$
Hence, mole fraction of glucose
$\qquad\qquad\qquad$$\begin{array}{l}=\frac{\text { Moles of glucose }}{\text { Total moles of solution }} \\ =\frac{10 / 180}{5+10 / 180}=\frac{0.055}{5+0.055}\end{array}$
Mole fraction of glucose $=\frac{0.055}{5.055}=0.01$
Therefore, mole fraction of water $=1-0.01=0.99$
(iii) Molarity (M) = $\frac{\text { Mass } \% \times \text { density } \times 10}{\text { Molar mass }}$
$\qquad\qquad\qquad$$M=\frac{10 \times 1.2 \times 10}{180}=0.666$
Hence, Molarity of the solution = 0.67

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Question 24 Marks

Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄ in 2 litre of water at 25°C, assuming that it is completely dissociated.

Answer

Complete dissociation of K₂SO₄,
$K _2 SO _4 \xrightarrow{ H _2 O } 2 K^{+}+ SO _4^{2-}$
Hence number of ions obtained from ionization (i) =3
$\text { Osmotic pressure }(\Pi)=\text { iCRT }$
$ R =0.0821$ $L$ $atm$ $mol ^{-1} K^{-1}, $
$T=25^{\circ} C +273.15=298.15, C = mol / \text { litre }$
$=\frac{\text { Mass of solute (g) }}{\text { Molar mass of solute } \times \text { Volume of solution (L) }}$
Mass of solute = 25 mg = 25 × 10^-3 g
$\text{Molar mass of } K _2 SO _4 =(2 \times 39)+32+(4 \times 16) $
$ =78+32+64=174$
Hence, concentration $(C) =\frac{25 \times 10^{-3}(g)}{174 \times 2(L)} $
$ =7.18 \times 10^{-5}$
Osmotic pressure $(\Pi)= iCRT$
$=3 \times\left(7.18 \times 10^{-5}\right) \times 0.0821 \times 298.15 $
$=527.25 \times 10^{-5}$
Hence, osmotic pressure of solution
$=5.27 \times 10^{-5} atm$.

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Question 34 Marks

Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.

Answer

Vapour pressure of water $p_1^0=17.535$ $mm$ $Hg$
Let vapour pressure of solution = p₁
Relative depression in vapour pressure
$=\frac{p_1^0-p_1}{p_1^0}=x_2$
n₁ = Moles of solvent (water) = $\frac{450}{18}$
Molar mass of glucose $\left( C _6 H _{12} O _6\right)=180$
$n_2=$ Moles of solute = $\frac{25}{180}$
$\begin{array}{l}x_2=\frac{p_1^0-p_1}{p_1^0} \\ x_2=1-\frac{p_1}{p_1^0}\end{array}$
$\begin{aligned} 1-x_2 & =\frac{p_1}{p_1^0} \text { or } p_1=p_1^0\left(1-x_2\right) \\ p_1 & =p_1^0 \times x_1, \quad\left(1-x_2=x_1\right)\end{aligned}$
$\begin{array}{l}p_1=17.535 \times \frac{450 / 18}{\frac{25}{180}+\frac{450}{80}} \\ p_1=17.535 \times \frac{25}{0.1388+25}\end{array}$
$\begin{array}{l}p_1=\frac{17.535 \times 25}{25.1388} \\ p_1=17.438 mm Hg \end{array}$
Hence, vapour pressure of solution = 17.44 mm Hg

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Question 44 Marks

A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.

Answer

Freezing point $\left(\Delta T_f\right)=\frac{K_f \times w_2 \times 1000}{M_2 \times w_1}$
5% (mass) of sugar means 5 g sugar + 95 g water
$\qquad\qquad\qquad$$\Delta T _{ f }=273.15-271=2.15$
Mass of sugar = 5 g = W₂, W₁= 95 g
Molar mass of sugar $\left( C _{12} H _{22} O _{11}\right) M _2=342$
$ K _{ f }=? $
$K_{ f }=\frac{\Delta T _{ f } \times M _2 \times w _1}{ w _2 \times 1000}$
On putting value
$K_f=\frac{2.15 \times 342 \times 95}{5 \times 1000} $
$K_f=13.97$
For 5% (mass) aqueous solution of glucose,
$\Delta T _{ f }=\frac{ K _{ f } \times w_2 \times 1000}{ M _2 \times w_1} $
$w_1=95 g, w_2=5 g$
M₂ = Molar mass of glucose $\left( C _6 H _{12} O _6\right)=180$
Therefore, $\Delta T _{ f }=\frac{13.97 \times 5 \times 1000}{180 \times 95} $
$\Delta T_{ f }=4.084$
Hence, freezing point of aqueous solution glucose
= Freezing point of water - Freezing point depression
=273.15 - 4.08 = 269.07
Hence, Freezing point of aqueous solution of glucose
= 269.07 K

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Question 54 Marks

Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?

Answer

Molar mass of heptane (C₇H₁₆)
= (7 × 12) + 16 = 100
Molar mass of octane (C₈H₁₈)
= (8 × 12) + 18 = 114
Mole of heptane (n₁)$=\frac{26}{100}=0.26$
Mole of octane (n₂) $=\frac{35}{114}=0.307$
Mole fraction of heptane (x₁) $=\frac{0.26}{0.26+0.307}$
$x_1=\frac{0.26}{0.567}=0.458$
Mole fraction of octane (x₂) $=1-x_1$
$=1-0.458=0.542$
Vapour pressure of heptane in solution $\left(p_1\right)=p_1^0 x_1$
$\begin{array}{l}=105.2 \times 0.458\left(p_1^0=105.2 kPa \right) \\ =48.18 kPa \end{array}$
Vapour pressure of octane in solution $= p _2^0 x _2$
$\begin{aligned} p_2 & =46.8 \times 0.542 \quad\left(\because p_2^0=46.8 kPa \right) \\ & =25.36 kPa \end{aligned}$
Vapour pressure of whole mixture $p=p_1+p_2$
$\begin{aligned} p & =48.18+25.36 \\ & =73.54 kPa \end{aligned}$

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