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, \[\frac{{{W}_{(Glu\cos e)}}\times 18}{50\times 180}=\frac{{{W}_{(urea)}}\times 18}{50\times 60}\] \[{{W}_{(Glu\cos e)}}=weight\,of\,glu\cos e\] \[{{W}_{(Glu\cos e)}}=weight\,of\,urea\] Or \[\frac{{{W}_{(Glu\cos e)}}\times 18}{50\times 180}=\frac{1\times 18}{50\times 60}\] \[{{W}_{(Glu\cos e)}}=3\]
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