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Standard XII

Physics

Area and Volume Expansion

The elements ...

Question

The elements $^{'} X^{'}$ and $^{'} Y^{'}$ form two different binary compounds $X Y_{2}$ and $X Y_{4}$ . When dissolved in $20 g m$ of $C S_{2}$ solvent, $1 g m$ of $X Y_{2}$ lowers the freezing point by ${2.5}^{o} C$ , whereas $1 g m$ of $X Y_{4}$ lowers the freezing point by ${1.5}^{o} C$ . The molal depression constant for $C S_{2}$ is $5$ . The atomic mass of element $^{'} Y^{'}$ will be :

42.64

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64.42

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33.34

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12.8

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Solution

The correct option is C $33.34$
$Δ T_{f_{1}} = - k_{f} m_{1}$
$2.5 = 5 \times \frac{1}{(M . w t)} \times \frac{1000}{20}$
$(M . w t)_{1} = 100$
$X + 2 Y = 100 - - (1)$
$Δ T_{f_{2}} = k_{f} m_{2}$
$1.5 = 5 \times \frac{1}{{(M . w t)}_{2}} \times \frac{1000}{20}$
${(M . w t)}_{2} = 166.66$
$X + 4 Y = 166.66 - - (2)$
$N o w (2) - (1)$
$2 Y = 66.66$
$Y = 33.33$

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QUESTION 2.21

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Two elements A and B form compounds of formula AB₂ and AB₄. When dissolved in 20.0g of benzene 1.0g of AB₂lowers freezing point by 2.3^°C whereas 1.0g of AB₄ lowers freezing point by 1.3^°C. The K_f for benzene. The atomic masses of A and B are

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The correct option is C 33.34ΔTf1=−kfm12.5=5×1(M.wt)×100020(M.wt)1=100X+2Y=100−−(1)ΔTf2=kfm21.5=5×1(M.wt)2×100020(M.wt)2=166.66X+4Y=166.66−−(2)Now(2)−(1)2Y=66.66Y=33.33