A) 1 g
B) 3 g
C) 6g
D) 18 g
Correct Answer: B
Solution :
\[\frac{p-{{p}_{s}}}{p}=\frac{{{w}_{1}}{{M}_{2}}}{{{w}_{2}}{{M}_{1}}}\] To produce same lowering of vapour pressure,\[\frac{p-{{p}_{s}}}{p}\]will be same for both cases. So, \[\frac{{{W}_{(Glu\cos e)}}\times 18}{50\times 180}=\frac{{{W}_{(Urea)}}\times 18}{50\times 60}\] \[{{W}_{(Glu\cos e)}}=weight\text{ }of\text{ }glu\cos e\] \[{{W}_{(Urea)}}=weight\text{ }of\text{ }glu\cos e\] Or \[\frac{{{W}_{(Glu\cos e)}}\times 18}{50\times 180}=\frac{1\times 18}{50\times 60}\] \[{{W}_{(Glu\cos e)}}=3g\]
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