Question

The molality of glucose when 0.015 moles of glucose dissolve in 30g of water .a) 0.50mb) 5.4m c) 1md) 0.3m​

Answer 1

Given :

➾ Moles of Glucose = 0.015

➾ Weight of water = 30g

To Find :

➠ Molality of the solution.

Molality :

➝ Symbol : m

➝ Definition : Number of moles of solute per 1kg of the solvent.

➝ Effect of temperature : No

➝ Unit : mol/kg

➝ Formula :

$\bigstar\:\underline{\boxed{\bf{\purple{m=\dfrac{Moles\:of\:solute}{Mass\:of\:solvent\:in\:kg}}}}}$

Conversion :

➳ 1000g = 1kg

➳ 30g = 30/1000 = 0.03

Calculation :

⇒ m = n / w

⇒ m = 0.015 / 0.03

⇒ m = 5/10

⇒ m = 0.5 mol/kg

Answer 2

Given:

◑Moles of glucose :- 0.015

◑Water :- 30g

To Find:

☞Molality of glucose.

Solution:

First we convert "g" into "kg".

➡1000g = 1kg

➡30g = 30/1000

➡0.03kg

Calculation:

⇒ m = n / w

⇒ m = 0.015 / 0.03

⇒ m = 5/10

⇒ m = 0.5mol/kg

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More to know:

• Molality - It is defined as the number of moles of solute per kilogram of the solvent.
• Molality (m) = No. of moles of solute/ weight of solvent (in kg).
• If number of solute is w and weight of solvent Is m then the formula of molality is "m = w / M x 1 / W (in Kg)".
• SI unit is moles per kilogram.

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