250g solution of D -glucose in water contains 10 . 8 % of carbon by...

${C}_{6}{H}_{12}{O}_{6} \to Glucose$

Given: $$

$\dfrac{10 . 8 \times 250}{100} =$ mass of $C ⇒$ Mass of $C = 27gm$

$\dfrac{mass of C}{massofglucose} = \dfrac{72}{180}massofglucose = 27 \times \dfrac{180}{72}$

$∴$ mass of glucose $= 67 . 5gm$

$∴$ moles of glucose $= \dfrac{67 . 5}{180} = 0 . 375$ moles

Mass of solvent $= 250 - 67 . 5gm = 182 . 5gm$

$∴$ Molality $= \dfrac{molesofsolute}{massofsolvent ( kg )} = \dfrac{0 . 375}{0 . 1825} = 2 . 055 \approx 2 . 06$