MCQ
A sample of a hydrate of barium chloride weighing $61\,g$ was heated until all the water of hydration is removed. The dried sample weight is $52\,g.$ The formula of the hydrated salt is : ( atomic mass, $Ba = 137\,amu,\,\,Cl = 35.5\,amu$ )
  • A
    $BaC{l_2}.4{H_2}O$
  • B
    $BaC{l_2}.3{H_2}O$
  • C
    $BaC{l_2}.{H_2}O$
  • $BaC{l_2}.2{H_2}O$

Answer

Correct option: D.
$BaC{l_2}.2{H_2}O$
d
Weight of hydrated $BaCl_2 = 61\,g$

Weight of anhydrous $BaCl_2 = 52\,g$

Loss in mass $= 9\,g$

Assuming $BaCl_2.xH_2O$ as hydrate mass of $H_2O = 9\,g$

Moles of ${{H}_{2}}O=\frac{9}{18}=0.5$

Grass molecular let of $BaCl_2 = 208$

$\%$ of $H_2O$ in this hydrated $BaC{{l}_{2}}=\frac{9}{61}\times 100$

$=14.75\%$

$=\frac{18x}{208+18x}\times 100$ on solving $x = 2$

This percentage is present in $BaCl_2.2H_2O$

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