# chemistry

A 43 mL sample of a solution of sulfuric acid is neutralized by 24 mL of a 0.053 M sodium hydroxide solution. Calculate the molarity of the sulfuric acid solution.

Here is my work so far. I thought I have successfully solved the problem, but apparently my answer Is wrong.

H2SO4 + 2NaOH = 2H2O + Na2SO4

.024L NaOH x .053M NaOH = .001272 mol NaOH
.001272 mol NaOH x (1 mol H2SO4/2 mol NaOH) / .043L H2SO4
= .0147906977

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1. Final answer is molarity of the H2SO4 btw.

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2. What makes you think it's wrong?

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3. The website we use for our homework tells you when your answer is wrong right when you type it in, and mine is apparently wrong.

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4. You answer of 0,0147906,,,,, should be rounded to 0.015, You may not have rounded. You may not have added M or mols/L ater the 0.015. Most of these auto checking websites are unforgiving on significant figures. Try the 0.015 and if that doesn't work bry 0.015 M.

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6. Try 0.030 as an answer. Maybe the website wants to know what the polarity is if you were just titrating to obtain the first equivalent point. I agree with 0.015, but 0.030 is worth a shot.

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