A) methanol
B) glucose
C) carbon dioxide
D) Cannot predict
Correct Answer: B
Solution :
\[\frac{p{}^\circ -{{p}_{s}}}{p{}^\circ }=\frac{n}{n+N}=\frac{\frac{w}{m}}{\frac{w}{m}+\frac{W}{M}}\] \[\because \] \[\frac{w}{m}<\frac{W}{M}\] \[\therefore \] \[\frac{p{}^\circ -{{p}_{s}}}{p{}^\circ }=\frac{w\text{/}m}{W\text{/}M}=\frac{w}{m}\times \frac{M}{W}\] \[\frac{3000-2985}{3000}=\frac{5}{m}\times \frac{18}{100}\] \[m=\frac{5\times 18\times 3000}{15\times 100}\] \[m=180\] 180 is the molecular weight of glucose\[({{C}_{6}}{{H}_{12}}{{O}_{6}}),\] thus the substance\[X\]is glucose.
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