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3 years ago

What is the concentration of a solution that has a volume of 2.5L and contains exactly 2.125 moles of calcium phosphate

Chemistry

1 answer:

Iteru [2.4K]3 years ago

7 0

Answer:

0.85 mol/L.

Explanation:

Molarity is defined as the no. of moles of solute dissolved in a liter of the solution.

M = (no. of moles of solute)/(Volume of the solution (L))

no. of moles of calcium phosphate = 2.125 mol.

Volume of the solution = 2.5 L.

∴ M of calcium phosphate = (2.125 mol)/(2.5 L) = 0.85 mol/L.

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A gas of unknown identity diffuses at a rate of 155 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate o

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19.07 g mol^-1

Explanation:

The computation of the molecular mass of the unknown gas is shown below:

As we know that

$\frac{Diffusion\ rate\ of unknown\ gas }{CO_{2}\ diffusion\ rate} = \frac{\sqrt{CO_{2\ molar\ mass}} }{\sqrt{Unknown\ gas\ molercular\ mass } }$

where,

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Now placing these values to the above formula

$\frac{155mL/s}{102mL/s} = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ 1.519 = \frac{\sqrt{44 g mol^{-1}} }{\sqrt{M_{unknown}} } \\\\ {\sqrt{M_{unknown}} } = \frac{\sqrt{44 g mol^{-1}}}{1.519} \\\\ {\sqrt{M_{unknown}} } = \frac{44 g mol^{-1}}{(1.519)^{2}}$

After solving this, the molecular mass of the unknown gas is

= 19.07 g mol^-1

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