How many moles of NaOH are present in 26,5 mL of 0.190 M NaOH?

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How many moles of NaOH are present in 26,5 mL of 0.190 M NaOH?

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Answer 1 · 2018-04-28T12:38:47

$0.00504 mol$ $"0.00504 mol"$ .

Explanation:

$0.190 M$ $"0.190 M"$ $N a O H$ $NaOH$ means that, for every $1$ $1$ liter of $N a O H$ $NaOH$ , there will be $0.190$ $0.190$ moles:

$\frac{0.190 mol}{1L}$ $"0.190 mol"/"1L"$

So, using this ratio, we can find how many moles of $N a O H$ $NaOH$ there are in $26.5 mL$ $"26.5 mL"$ , or $0.0265 L$ $"0.0265 L"$ :

$\frac{0.190 mol}{1L} = \frac{0.190 \times 0.0265 mol}{0.0265 L} = \frac{0.00504 mol}{0.0265 L}$ $"0.190 mol"/"1L" = "0.190 × 0.0265 mol"/"0.0265 L" = "0.00504 mol"/"0.0265 L"$

Answer 2 · 2018-04-28T12:45:14

0.00504

Explanation:

Concentration = moles of solute / volume of solution.

$sf(c=n/v)$

$:.$ $sf(n=cxxv)$

$:.$ $sf(n_(NaOH)=0.0190xx26.5/1000=0.00504)$

2 points to note:

I have converted ml to litre by dividing by 1000.
I have rounded to 3 sig. figs since the volume is given to this significance.

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How many moles of NaOH are present in 26,5 mL of 0.190 M NaOH?

2 Answers

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