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3 years ago

Find the percent of water in Zn3(PO4)2 x 7H2O

Chemistry

1 answer:

Andreas93 [3]3 years ago

4 0

Answer:

24.66%

Explanation:

To find the percent of water (H2O) in the following compound: Zn3(PO4)2. 7H2O

The atomic mass of the elements are as follows:

Zn = 65g/mol

P = 31g/mol

O = 16g/mol

H = 1g/mol

Hence, molar mass of the compound is as follows:

65(3) + {31 + 16(4)} 2 + 7{1(2) + 16}

195 + 95(2) + 7(18)

195 + 190 + 126

511g/mol

Molar mass of 7H2O

= 7(18)

= 126g/mol

Hence, the percent of H2O in the compound is:

Molar Mass of 7H2O/molar mass of compound × 100

= 126/511 × 100

= 0.2466 × 100

= 24.66%

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3 years ago

The mole fraction of a non-electrolyte (MM 40.0 g/mol) in a saturated aqueous solution is 0.310. What is the molality of the sol

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Answer: The molality of non-electrolyte is 24.69 m

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We are given:

Mole fraction of saturated aqueous solution = 0.310

This means that 0.310 moles of non-electrolyte is present.

Moles of water (solvent) = 1 - 0.310 = 0.690 moles

To calculate the mass from given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of water = 0.690 moles

Molar mass of water = 18 g/mol

Putting values in above equation, we get:

$0.690mol=\frac{\text{Mass of water}}{18g/mol}\\\\\text{Mass of water}=(0.690mol\times 18g/mol)=12.42g$

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{n_{solute}\times 1000}{W_{solvent}\text{ (in grams)}}$

Where,

$n_{solute}$ = Moles of solute (non-electrolyte) = 0.310 moles

$W_{solvent}$ = Mass of solvent (water) = 12.42 g

Putting values in above equation, we get:

$\text{Molality of non-electrolyte}=\frac{0.310\times 1000}{12.42}\\\\\text{Molality of non-electrolyte}=24.96m$

Hence, the molality of non-electrolyte is 24.69 m

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