Question

18g glucose and 6g urea are dissolved in 1l of solutions fin d molarity

Answer 1

Molarity = 4.926 atm

Explanation:

Given,

The weight of gucose$(w_{1})=18g$ and

The weight of gucose$(w_{2})=6g$

To find, molarity = ?

The molecular mass of gocose,

$C_{6}H_{12}O_{6}=6\times 12+12\times 1+6\times 16<strong>=180$

The molecular mass of urea,

$CO(NH_{2})_{2}=12+16+(14+2)\times 2<strong>=60$

We know that,

$\pi V=(n_{1}+n_{1})RT$

⇒ $\pi V=(\dfrac{w_{1}}{m_{1}}+\dfrac{w_{1}}{m_{1}})RT$

⇒ $\pi \times 1=(\dfrac{18}{180}+\dfrac{6}{60}}})\times 0.0821 \times 300$

⇒ $\pi=(10+0.10)\times 0.0821 \times 300$

⇒ $\pi=4.926atm$

Hence, molarity = 4.926 atm

Answer 2

Molarity of given solution is - 0.2 M

Molarity is defined as "the concentration of a chemical species, in particular of a solute in a solution, in terms of amount of substance per unit volume of solution."

Formula to calculate molarity is -

M = number of moles/ volume of solution in litre.

Number of moles is calculated by the formula-

Weight/ molecular weight

Given in question -

Weight of glucose - 18g

Weight of urea - 6g

Volume of solution - 1 lt

Molecular weight of glucose -

C6H12O6 - 180 g

Molecular weight of urea -

CH₄N₂O - 60g

Number of moles of solute - (18/180)+(6/60)

= 0.1+0.1

Number of moles of solute = 0.2

Therefore, molarity = 0.2/1

Thus, molarity of given solution is = 0.2 M