# What is the percent composition of water in #Na_2S • 9H_2O?#

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To find the percent composition of water in Na₂S • 9H₂O, first calculate the molar mass of water and the total molar mass of the compound. Then divide the molar mass of water by the total molar mass of the compound and multiply by 100 to get the percent composition.

Molar mass of water (H₂O) = 2(1.008 g/mol) + 1(15.999 g/mol) = 18.015 g/mol

Total molar mass of Na₂S • 9H₂O: Na: 2(22.99 g/mol) = 45.98 g/mol S: 1(32.07 g/mol) = 32.07 g/mol H₂O: 9(18.015 g/mol) = 162.135 g/mol

Total molar mass = 45.98 g/mol + 32.07 g/mol + 162.135 g/mol = 240.185 g/mol

Percent composition of water: (162.135 g/mol ÷ 240.185 g/mol) × 100 = 67.5%

Therefore, the percent composition of water in Na₂S • 9H₂O is 67.5%.

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To find the percent composition of water in Na2S • 9H2O, first calculate the molar mass of water (H2O) and the molar mass of the entire compound (Na2S • 9H2O). Then, divide the molar mass of water by the molar mass of the compound and multiply by 100 to obtain the percent composition.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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