Question

How many grams of beryllium chloride would you need to add to make 250 mL of 0.05 molar solution?

Answer 1

Given: Molarity, M=0.05M, Volume, V=250mL, M.M. of BeCl₂=80g/mol

Formula:

• Molarity=number of moles of solute/Volume of solution(Litres)
• Number of moles=given mass/molar mass

⇒M=(m×1000)÷[V(mL)×M.M]

★Solution★

$0.05mol \: l {}^{ - 1} = \frac{m \times 1000}{80 \times 250}$

$m = \frac{0.05 \times 80 \times250 }{1000}$

⇒m=1g

Answer 2

Answer:

1 g

Explanation:

Given :

Volume = 250 mL

Molarity = 0.05 M

We are asked to find given mass of beryllium chloride BeCl₂ :

We have molar mass of beryllium chloride 79.91 ≈ 80 g / mol

We know :

Molarity = Given mass / ( Volume ( mL ) × Molar mass ) × 1000

Given mass = Molarity × ( Volume ( mL ) × Molar mass ) / 1000

Putting values here we get :

Given mass = ( 0.05 × 250 × 80 ) / 1000 g

= > Given mass = ( 0.05 × 80 ) / 4 g

= > Given mass = ( 4 ) / 4 g

= > Given mass =  1 g

Hence we get required answer.