Borax in water gives B4O72- + 7H2O→ 4H3BO3 + 2OH How many grams of Borax (Na2B4O7 . 10H2O) are required to prepare 50 ml of 0.2 M solution.

Question

Borax in water gives B4O72- + 7H2O→ 4H3BO3 + 2OH How many grams of Borax (Na2B4O7 . 10H2O) are required to prepare 50 ml of 0.2 M solution. (A) 0.32 g (B) 3.82 g (C) 0.28 g (D) None of these

Tardigrade Admin · Accepted Answer

The n-factor or valency factor of borax in the above reaction is 2 Meq. of borax in 50 ml solution = 50 × 0.2 × 2 = 20 Let weight of borax required = W (W/382) × 2 × 1000 = 20 ∴ W(Na2B4O7. 10H2O) = 3.82

Q. Borax in water gives
$B_{4} O_{7}^{2 -} + 7 H_{2} O \to 4 H_{3} B O_{3} + 2 O H$
How many grams of Borax $(N a_{2} B_{4} O_{7} .10 H_{2} O)$ are required to prepare $50 m l$ of $0.2 M$ solution.

$0.32 g$

$3.82 g$

$0.28 g$

None of these

The $n$ -factor or valency factor of borax in the above reaction is $2$
Meq. of borax in $50 m l$ solution $= 50 \times 0.2 \times 2 = 20$
Let weight of borax required $= W$
$\frac{W}{382} \times 2 \times 1000 = 20$
$∴ W_{(N a_{2} B_{4} O_{7} .10 H_{2} O)} = 3.82$

Q. Borax in water gives B4​O72−​+7H2​O→4H3​BO3​+2OH How many grams of Borax (Na2​B4​O7​.10H2​O) are required to prepare 50ml of 0.2M solution.

0.32g

3.82g

0.28g