The mole fraction of glucose ( C 6 H 12 O 6) in an aqueous binary solution is 0.1 . The mass percentage of water in it, to the nearest integer, is.

Question

The mole fraction of glucose ( C 6 H 12 O 6) in an aqueous binary solution is 0.1 . The mass percentage of water in it, to the nearest integer, is. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

X C 6 H 12 O 6=0.1 Let total mole is 1 mol then mole of glucose will be 0.1 and mole of water will be 0.9 so mass % of water =(0.9 × 18/0.1 × 180+0.9 × 18) × 100 = 47.36

Q. The mole fraction of glucose $(C_{6} H_{12} O_{6})$ in an aqueous binary solution is $0.1.$ The mass percentage of water in it, to the nearest integer, is_________.

$X_{C_{6} H_{12} O_{6}} = 0.1$

Let total mole is $1$ mol then mole of glucose will be $0.1$ and mole of water will be $0.9$

so mass $%$ of water $= \frac{0.9 \times 18}{0.1 \times 180 + 0.9 \times 18} \times 100$

$= 47.36$

Q. The mole fraction of glucose (C6​H12​O6​) in an aqueous binary solution is 0.1. The mass percentage of water in it, to the nearest integer, is_________.

XC6​H12​O6​​=0.1 Let total mole is 1 mol then mole of glucose will be 0.1 and mole of water will be 0.9 so mass % of water =0.1×180+0.9×180.9×18​×100 =47.36

Q. The mole fraction of glucose $(C_{6} H_{12} O_{6})$ in an aqueous binary solution is $0.1.$ The mass percentage of water in it, to the nearest integer, is_________.

$X_{C_{6} H_{12} O_{6}} = 0.1$

Let total mole is $1$ mol then mole of glucose will be $0.1$ and mole of water will be $0.9$

so mass $%$ of water $= \frac{0.9 \times 18}{0.1 \times 180 + 0.9 \times 18} \times 100$

$= 47.36$