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pickupchik [31]
3 years ago
12

What volume of solution must be added to 4.0 mol of NaCl to make a 1.2 M solution?

Chemistry
1 answer:
Aleksandr [31]3 years ago
7 0

Answer:

\boxed {\boxed {\sf 3.3 \ liters}}}

Explanation:

Molarity is a measure of concentration in moles per liter.

molarity=\frac{moles \ of \ solute}{liters \ of \ solution}}

The solution has a molarity of 1.2 M or 1.2 moles per liter. There are 4.0 moles of NaCl, the solute. We don't know the liters of solution, so we can use x.

  • molarity= 1.2 mol/L
  • moles of solute= 4.0 mol
  • liters of solution =x

Substitute the values into the formula.

1.2 \ mol/L = \frac{4.0 \ mol}{x}

Since we are solving for x, we must isolate the variable. Begin by cross multiply (multiply the 1st numerator and 2nd denominator, then the 1st denominator and 2nd numerator.

\frac {1.2 \ mol/L}{1}=\frac{ 4.0 \ mol}{x}

4.0 \ mol *1=1.2 \ mol/L *x

4.0 \ mol = 1.2 \ mol/L *x

x is being multiplied by 1.2 moles per liter. The inverse of multiplication is division, so divide both sides by 1.2 mol/L

\frac{4.0 \ mol}{1.2 \ mol/L} = \frac{1.2 \ mol/L *x}{1.2 \ mol/L}

\frac{4.0 \ mol}{1.2 \ mol/L}=x

The units of moles (mol) will cancel.

\frac{4.0 }{1.2 } \ L =x

3.33333333 \ L=x

The original measurements both have 2 significant figures, so our answer must have the same. For the number we found, this is the tenths place.

The 3 in the hundredth place tells us to leave the 3 in the tenths place.

3.3 \ L\approx x

Approximately  <u>3.3 liters of solution</u> are needed to make a 1.2 M solution with 4.0 moles of sodium chloride.

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When 0.513 g of biphenyl (c12h10 undergoes combustion in a bomb calorimeter, the temperature rises from 26.3 ?c to 29.7 ?c?
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When 0.514 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.8 C to 29.4 C. Find ⌂E rxn for the combustion of biphenyl in kJ/mol biphenyl. The heat capacity of the bomb calorimeter, determined in a separate experiment, is 5.86 kJ/ C. 



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The equilibrium constant, Kc, for the following reaction is 5.10×10-6 at 548 K. NH4Cl(s) NH3(g) + HCl(g) Calculate the equilibri
levacccp [35]

<u>Answer:</u> The equilibrium concentration of HCl is 2.26\times 10^{-3}M

<u>Explanation:</u>

We are given:

Moles of NH_4Cl(s) = 0.564 moles

Volume of vessel = 1.00 L

Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}

Molarity of NH_4Cl=\frac{0.564}{1}=0.564M

The given chemical equation follows:

                  NH_4Cl(s)\rightleftharpoons NH_3(g)+HCl(g)

<u>Initial:</u>         0.564

<u>At eqllm:</u>     0.564-x          x              x

The expression of K_c for above equation follows:

K_c=[NH_3][HCl]

The concentration of pure solid and pure liquid is taken as 1.

We are given:

K_c=5.10\times 10^{-6}

Putting values in above equation, we get:

5.10\times 10^{-6}=x\times x\\\\x=2.26\times 10^{-3}M,-2.26\times 10^{-3}M

Negative sign is neglected because concentration cannot be negative.

So, [HCl]=2.26\times 10^{-3}M

Hence, the equilibrium concentration of HCl is 2.26\times 10^{-3}M

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3 years ago
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