A) 1 g
B) 3 g
C) 6 g
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, \[\frac{{{W}_{(Glucose)}}\times 18}{50\times 18}=\frac{{{W}_{(urea)}}\times 18}{50\times 60}\] \[{{W}_{(Glucose)}}\]= weight of glucose \[{{W}_{(Glucose)}}\]= weight of urea \[\frac{{{W}_{(Glucose)}}\times 18}{50\times 18}=\frac{1\times 18}{50\times 60}\] \[{{W}_{(Glucose)}}=3\]
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