A) \[1g\]
B) \[3g\]
C) \[6g\]
D) \[18g\]
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, \[\frac{{{W}_{(Glu\cos e)}}\times 18}{50\times 180}=\frac{{{W}_{(turea)}}\times 18}{50\times 60}\] \[{{W}_{(Glucose)}}\]=weight of glucose \[{{W}_{(Glucose)}}\] = weight of urea or \[\frac{{{W}_{(Glucose)}}\times 18}{50\times 180}=\frac{1\times 18}{50\times 60}\] \[{{W}_{(Glucose)}}=3\]
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