# What volume of 0.123 M NaOH in milliliters contains 25.0 g NaOH?

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As I understand it:

I gather that NaOH has a molar mass of about 40g/mol.

If the concentration of NaOH is 0.123 mol/L then I have 0.492 g/L which is the same as 0.492 mg/mL.

So (0.492mg/1mL)(25000mg) should equal 50813 mL. But the correct answer is 5080 mL according to my book. Where did I go wrong?

As I understand it:

I gather that NaOH has a molar mass of about 40g/mol.

If the concentration of NaOH is 0.123 mol/L then I have 0.492 g/L which is the same as 0.492 mg/mL.

So (0.492mg/1mL)(25000mg) should equal 50813 mL. But the correct answer is 5080 mL according to my book. Where did I go wrong?

##### 1 Answer

#### Answer:

#### Explanation:

Your book is rounding the answer to three **significant figures**.

Your calculations are actually off by an order of magnitude because

#0.123 color(red)(cancel(color(black)("moles NaOH")))/"L" * "40 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = "4.92 g/L"#

not

#"4.92 g/L " = " 4.92 mg/mL"#

and

#25000color(red)(cancel(color(black)("mg"))) * "1 mL"/(4.92color(red)(cancel(color(black)("mg")))) = "5081.3 mL"#

The answer is indeed

In this case, the values given for the molarity of the solution and the mass of sodium hydroxide are rounded to three sig figs, which is why the answer **must** have three sig figs.

Think of it like this -- the answer **cannot** be more precise than the *least precise* measurement.

Also, you must use a value for the **molar mass** of sodium hydroxide that is consistent with the number of sig figs you have for the measurements.

Unless the problem explicitly mentions this, you should use

I'll use a series of *conversion factors* to get the answer

#25.0 color(red)(cancel(color(black)("g"))) xx overbrace((1color(red)(cancel(color(black)("mole NaOH"))))/(39.997color(red)(cancel(color(black)("g")))))^(color(blue)("molar mass of NaOH")) xx overbrace((1color(red)(cancel(color(black)("L"))))/(0.123color(red)(cancel(color(black)("moles NaOH")))))^(color(purple)("the given molarity"))#

# xx overbrace((10^3"mL")/(1color(red)(cancel(color(black)("L")))))^(color(darkgreen)("L to mL")) = "5081.7 mL"#

Rounded to **three sig figs**, the answer will indeed be

#color(green)(bar(ul(|color(white)(a/a)color(black)("volume in mL " = " 5080 mL")color(white)(a/a)|)))#

So remember, *always* double-check your calculations and keep track of your significant figures!