By usual analysis, 1.00g of compound (X) gave 1.79g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is : ______ (nearest integer) (Given, molar mass in g : O = 16, Mg = 24, P = 31) [2026]

Question

By usual analysis, 1.00 g of compound (X) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is : ______ (nearest integer)

(Given, molar mass in g ${m o l}^{- 1}$ : O = 16, Mg = 24, P = 31) [2026]

Smart Achievers · Accepted Answer

(1)

${m o l}^{- 1}$

$= \frac{(\frac{1.79}{222} \times 2 \times 31)}{1} \times 100$

$= 49.99 % \approx 50 % .$

Solution

Q.
By usual analysis, 1.00 g of compound (X) gave 1.79 g of magnesium pyrophosphate. The percentage of phosphorus in compound (X) is : ______ (nearest integer)

(Given, molar mass in g ${m o l}^{- 1}$ : O = 16, Mg = 24, P = 31) [2026]

1 50

2 30

3 40

4 20

Ans.
(1)

$% of P = \frac{n_{{Mg}_{2} P_{2} O_{7}} \times 2 \times 31}{W_{(unknown compound)}} \times 100$

$= \frac{(\frac{1.79}{222} \times 2 \times 31)}{1} \times 100$

$= 49.99 % \approx 50 % .$

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