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3 years ago

What is the molar mass of 37.96 g of gas exerting a pressure of 3.29 on the walls of a 4.60 L container at 375 K?

Chemistry

1 answer:

Phantasy [73]3 years ago

8 0

The molar mass of the gas is 77.20 gm/mole.

Explanation:

The data given is:

P = 3.29 atm, V= 4.60 L T= 375 K mass of the gas = 37.96 grams

Using the ideal Gas Law will give the number of moles of the gas. The formula is

PV= nRT (where R = Universal Gas Constant 0.08206 L.atm/ K mole

Also number of moles is not given so applying the formula

n= mass ÷ molar mass of one mole of the gas.

n = m ÷ x ( x molar mass) ( m mass given)

Now putting the values in Ideal Gas Law equation

PV = m ÷ x RT

3.29 × 4.60 = 37.96/x × 0.08206 × 375

15.134 = 1168.1241 ÷ x

15.134x = 1168.1241

x = 1168.1241 ÷ 15.13

x = 77.20 gm/mol

If all the units in the formula are put will get cancel only grams/mole will be there. Molecular weight is given by gm/mole.

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$M({ \left Na \right }_{ 2 } { \left So \right }_{ 4 }) = m \left( Na \right) +m \left( S \right) +m \left( O \right)$

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$\mathrm{M =molar\:mass }$

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$\mathrm{Substitute\:the\:values\:into\:the\:formula}$

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