Question

A 12.0 g sample of HF is dissolved in water to give 3.1 \times 10^23.1×10 2 mL of solution. The concentration of the solution is

Answer 1

Answer:

1:93 m

Explanation:

We can start with the data given by the problem:

Mass of HF = 12.0 g

Volume of solution = 3.1x10^23.1x10

2

mL

If we want to calculate the "concentration", we have to remember the molarity equation:

M=\frac{mol}{L}M=

L

mol

So, we need to know the moles and the litters if we want to calculate the molarity.

Calculation of Litters

We already have a volume value, but the value given by the problem has "mL". So, we have to do a conversion. In 1 L we have 1000 mL (1 L = 1000 mL), therefore:

3.1x10^2~mL\frac{1~L}{1000~mL}=0.31~L3.1x10

2

mL

1000 mL

1 L

=0.31 L

Calculation of moles

In this case, the compound is HF. So, we need to know the molar mass of HF if we want to do the conversion from grams to moles. The atomic mass of "F" is 18.99 g/mol and the atomic mass of "H" is 1 g/mol. So:

18.99 + 1 = 20 g /mol

In other words, 1 mol of HF is equal to 20 g (1 mol = 20 g HF). With this in mind, we can calculate the moles in the 12 g:

12~g~HF\frac{1~mol~HF}{20~g~HF}=~0.6~mol~HF12 g HF

20 g HF

1 mol HF

= 0.6 mol HF

Calculation of the concentration of the solution

With these values know we can calculate the molarity:

M=\frac{0.6~mol~HF}{0.31~L}=1.93~MM=

0.31 L

0.6 mol HF

=1.93 M

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