$(Image)$
$x=$.. . . . . .(નજીક નો પૂર્ણાક)
[આપેલ : $273.15 \mathrm{~K}$ પર પાણીનો મોલલ ઠારણ બિંદુ અવનયન અયળાંક $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$ છે]
\(\Delta \mathrm{T}_{\mathrm{f}}=\mathrm{K}_{\mathrm{f}} \mathrm{m} \Rightarrow 2.5=1.86 \times \frac{\mathrm{n}}{0.1}\)
\(\Rightarrow \mathrm{n}=0.1344 \text { moles }\)
\(\Rightarrow \mathrm{w}=0.1344 \times 32=4.3 \mathrm{~g}\)
\(\text { Volume }=\frac{4.3}{0.792}=5.43 \mathrm{ml}=543 \times 10^{-2} \mathrm{ml}\)
$A.$ $0.500\,M\,C _2 H _5 OH ( aq )$ અને $0.25\, M\, KBr ( aq )$
$B.$ $0.100\,M\,K _4\left[ Fe ( CN )_6\right]$ (aq) અને $0.100\, M$ $FeSO _4\left( NH _4\right)_2 SO _4$ (aq)
$C.$ $0.05 \,M\, K _4\left[ Fe ( CN )_6\right]( aq )$ અને $0.25\, M\, NaCl$ (aq)
$D.$ $0.15\, M\, NaCl ( aq )$ અને $0.1\, M BaCl _2$ (aq)
$E.$ $0.02\, M\, KCl\, MgCl _{2 .} 6 H _2 O ( aq )$ અને $0.05\, M$ $KCl ( aq )$
(આણ્વિય દળ $\left.\mathrm{H}_{2} \mathrm{SO}_{4}=98 \;\mathrm{g} / \mathrm{mol}\right)$