The % of boron in borax ( $N a_{2} B_{4} O_{7} .10 H_{2} O$ ) is :
[ $H = 1, B = 11, O = 16, N a = 23$ ]

22.5 %

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17.4 %

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11.5 %

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23 %

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Solution

The correct option is D 11.5 %
Molar mass of $N a_{2} B_{4} O_{7} .10 H_{2} O$ $= 381 g$
So, % $B$ $= 44 \times \frac{100}{381} = 11.54$ %

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Similar questions

Calculate the percentage of Boron ( $B$ ) in $N a_{2} B_{4} O_{7} .10 H_{2} O$ . Answer correct to 1 decimal place. [H = 1, B = 11, O = 16, Na = 23].

Find the percentage of
(a) oxygen in magnesium nitrate crystals $[M g (N O_{3})_{2} 6 H_{2} O]$ .
(b) boron in $N a_{2} B_{4} O_{7} . 10 H_{2} O . [H = 1, B = 11, O = 16, N a = 23] .$
(c) phosphorus in the fertilizer superphosphate $C a (H_{2} P O_{4})_{2}$

(a) Calculate the percentage of boron (B) in borax (Na₂B₄O₇·10H₂O). (H = 1, B = 11, O = 16, Na = 23).
(b) (i) The compound A has the following percentage composition by mass:
Carbon = 26.7%, oxygen = 71.1%, hydrogen = 2.2%. Determine the empirical formula of A (H = 1, C = 12, O = 16)
(ii) If the relative molecular mass of A is 90, what is the molecular formula of A?

Q. What is the number of moles of sodium hydroxide contained in

160

g of it?

[N a = 23, O = 16, H = 1]

Q. Identify the type of change occurs in the following reaction:

_{11}^{23} N a +_{1}^{2} H ⟶_{11}^{24} N a +_{1}^{1} H

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The % of boron in borax (Na2B4O7.10H2O) is : [H=1,B=11,O=16,Na=23]

The correct option is D 11.5 %Molar mass of Na2B4O7.10H2O =381 gSo, % B =44×100381=11.54%

The % of boron in borax ( $N a_{2} B_{4} O_{7} .10 H_{2} O$ ) is :
[ $H = 1, B = 11, O = 16, N a = 23$ ]

The correct option is D 11.5 %
Molar mass of $N a_{2} B_{4} O_{7} .10 H_{2} O$ $= 381 g$
So, % $B$ $= 44 \times \frac{100}{381} = 11.54$ %