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deff fn [24]
3 years ago
14

What is the molarity of a solution made by dissolving 10.305 g of calcium carbonate, CaCO3, in water and diluting with water to

110.0 mL total?

Chemistry
1 answer:
Gemiola [76]3 years ago
8 0

Answer:

0.9359 M

Explanation:

M(CaCO3)=100.1 g/mol

10.305 g * 1 mol/100.1 g = 10.305/100.1 mol CaCO3

(10.305/100.1 mol )/0.1100 L = 0.936 mol/L = 0.9359 M

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Gas has a volume of 247.3 ML and is at 100 Celsius and 745 Hg. If the mass of the gas is 0.347 g what is the molar mass of the v
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Answer:

The molar mass of the vapor is 43.83 g/mol

Explanation:

Given volume of gas = V = 247.3 mL = 0.2473 L

Temperature = T = 100^{\circ}C = 373 K

Pressure of the gas = P = 745 mmHg  (1 atm = 760 mmHg)

P = \displaystyle \frac{745}{760} \textrm{ atm} = 0.9802 \textrm{ atm}

Mass of vapor = 0.347 g

Assuming molar mass of gas to be M g/mol

The ideal gas equation is shown below

\textrm{PV} =\textrm{nRT} \\\textrm{PV} = \displaystyle \frac{m}{M}\textrm{ RT } \\0.98026 \textrm{ atm}\times 0.2473 \textrm{ L} = \displaystyle \frac{3.47 \textrm{ g}}{M}\times 0.0821 \textrm{ L.atm.mol}^{-1}.K^{-1}\times 373\textrm{K} \\M = 43.834 \textrm{ g/mol}

The molar mass of the vapor comes out to be 43.834 g/mol

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