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

Question

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 (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

X textGlucose =0.1 mass % of glucose =(0 . 1 × 180/0 . 1 × 180 + 0 . 9 × 18)× 100 =(1800/18 + 16 . 2) =(1800/34 . 2) % =52.63 % =53 % ∴ mass % of H2O=47 %

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

$X_{Glucose} = 0.1$
mass $%$ of glucose $= \frac{0.1 \times 180}{0.1 \times 180 + 0.9 \times 18} \times 100$
$= \frac{1800}{18 + 16.2}$
$= \frac{1800}{34.2} %$
$= 52.63%$
$= 53%$
$∴$ mass $%$ of $H_{2} O = 47%$

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

XGlucose​=0.1 mass % of glucose =0.1×180+0.9×180.1×180​×100 =18+16.21800​ =34.21800​% =52.63% =53% ∴ mass % of H2​O=47%

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

$X_{Glucose} = 0.1$
mass $%$ of glucose $= \frac{0.1 \times 180}{0.1 \times 180 + 0.9 \times 18} \times 100$
$= \frac{1800}{18 + 16.2}$
$= \frac{1800}{34.2} %$
$= 52.63%$
$= 53%$
$∴$ mass $%$ of $H_{2} O = 47%$