Question

calculate the percentage of boron in borax,NA2B4O7.10H2O.

(H=1 B=11 O=16 Na=23)

(answer upto one decimal place)

please answer its very urgent

Answer 1

Answer:

11.5%

Explanation:

Na+B+O+H=(23*2)+(11*4)+(16*7)+(10*2)+16*10=382g

weight of boron in borax=(11*4)=44g

% of boron =(44/382)*100=11.5%

Answer 2

Given:

The given compound is $Na_2B_4O_7.10H_2O$.

To find:

The percentage of boron.

Solution:

The given compound is $Na_2B_4O_7.10H_2O$.

The atomic weight of Na is 23 grams.

The atomic weight of B is 11 grams and H is 1 gram.

The atomic weight of Oxygen is 16 grams.

Hence the molecular weight of borax is as follows-

$23 \times 2 + 11 \times 4 + 16 \times 7 + 10 \times (2 \times 1 + 16) \\ = 46 + 44 + 112 + 180 \\ = 382$

The amount of B present is 44 grams.

So, the percentage of B present is as follows-

$\frac{44}{382} \times 100 \\ = 11.5\%$

So, the percentage of B present in borax is 11.5%.