If 18 gram of glucose $(C_{6} H_{12} O_{6})$ is present in 1000 grams of an aqueous solution of glucose , it is said to be

1 m o l a l

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.1 m o l a l

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.5 m o l a l

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.1 m o l a l

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $0.1 m o l a l$
$Glucose givan mass=18g$
$Molar mass of glucose =180 g/mol$

we know that
$M o l a l i t y = \frac{mole of solute}{mass of solvent (k g)}$

$M o l a l i t y = \frac{18 / 180}{1000 / 1000} = 0.1 m o l a l$

correct answer is D.

Suggest Corrections

Similar questions

18 g of glucose (C₆H₁₂O₆) is present in 1000 g of an aqueous solution of glucose. The molality of this solution is:

1 k g

Molar mass of glucose is 180 g m o l^{- 1}

18 g

1000 g

100

(C_{6} H_{12} O_{6})

10 %

1

Join BYJU'S Learning Program