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djverab [1.8K]
3 years ago
13

A solution of H2SO4(aq) with a molal concentration of 4.80 m has a density of 1.249 g/mL. What is the molar concentration of thi

s solution?
Chemistry
1 answer:
Naddika [18.5K]3 years ago
5 0
Imagine we have <span>mass of solvent 1kg (1000g)
According to that: </span>molality =n(solute) / m(solvent) =\ \textgreater \  4.8m = 4.8mole / 1kg
= 4.8 mole * 98 g/mole = 470g
m(H2SO4) = n(H2SO4)*Mr(H2SO4) =\ \textgreater \=\ \textgreater \  Molarity = 4.8 mole / 1.2 L = 4 M m(H2SO4)  which is =<span>470g

</span><span>m(solution) = m(H2SO4) + m(solvent) = 470 + 1000 = 1470 g
d(solution) = m(solution) / V(solution) =>
=> 1.249 g/mL = 1470 g / V(solution) =></span>
Molarity = n(solute) / V(solution) =\ \textgreater \
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Part APart complete A sample of sodium reacts completely with 0.568 kg of chlorine, forming 936 g of sodium chloride. What mass
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Answer: 368 grams of sodium reacted.

Explanation:

The balanced reaction is :

2Na(s)+Cl_2(g)\rightarrow 2NaCl(s)

\text{Moles of chlorine}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of chlorine}=\frac{0.568\times 1000g}{71g/mol}=8moles

\text{Moles of sodium chloride}=\frac{936g}{58.5g/mol}=16mol    

According to stoichiometry :

2 moles of NaCl are formed from = 2 moles of Na

Thus 16 moles of NaCl are formed from=\frac{2}{2}\times 16=16moles  of Na

Mass of Na=moles\times {\text {Molar mass}}=16moles\times 23g/mol=368g

Thus 368 grams of sodium reacted.

4 0
3 years ago
2) Nitrogen gas will react with hydrogen gas to produce ammonia, NH3, How many
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Answer:

12 grams of hydrogen gas

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The molar mass of ammonia is 17 g/mol.

68 grams of ammonia corresponds to  

17g/mol

68g

​

=4moles

4 moles of ammonia will be obtained from  

2

4×1

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=2  moles of nitrogen and  

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4×3

​

=6  moles of hydrogen.

The molar masses of nitrogen and hydrogen are 28 g/mol and 2 g/mol respectively.

2 moles of nitrogen corresponds to 2×28=56  grams.

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2 years ago
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Question 15 How many grams of NaCl are required to make 500.0 mL of a 1.500 M solution? 58.40 g 175.3 g 14.60 g 43.83 g
ExtremeBDS [4]
Hi!

To make 500 mL of a 1,500 M solution of NaCl you'll require 43,83 g

To calculate that, you will need to use a conversion factor to go from the volume of the 1,500 M solution to the required grams. For this conversion factor, you'll use the definition for Molar concentration (M=mol/L) and the molar mass of NaCl. The conversion factor is shown below:

gNaCl=500mLsol* \frac{1L}{1000 mL}* \frac{1,500 mol NaCl}{1Lsol}* \frac{58,4428 g NaCl}{1 mol NaCl} \\ =43,83gNaCl

Have a nice day!
4 0
3 years ago
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