A sample of solid monoprotic acid with molar mass equal to 169.7 g/mol was titrated with 0.1599M sodium hydroxide solution. Calculate the mass in grms fo acid to be used if the volume of NAOH to be used is 25ml.
My calculation
0.1599 x 0.025L = 0.00399
0.00399 x 1000 = 3.9975
3.998 x 169.7 = 678.4 changes gm
x= 0.678gm
Except for the number of significant figures in your answer I agree with your answer but I don't understand exactly what you did.
First, since 0.1599 x 0.250 = 0.0039975, rounding to 0.00399 loses one significant figure. Multiplying by 1000 gets that back, I see, but 3.9975 actually should be 4. But why multiply by 1000. You have mols and that's what you need. And what is a "changes gm"? That could be milligram but then you rounded grams off to one fewer s.f. than is allowed. The answer is 0.67838 g which rounds to 0.6784 g (and all of this means I assumed the volume as 25.00 mL).
Your calculation is mostly correct, but there is a small error in converting the final mass to grams. Let's walk through the steps together:
1. Calculate the number of moles of sodium hydroxide (NaOH) used:
Moles of NaOH = molarity × volume of NaOH
Moles of NaOH = 0.1599 mol/L × 0.025 L = 0.0039975 mol
2. Determine the molar ratio between the acid and sodium hydroxide. Since the acid is monoprotic, the mole ratio is 1:1. This means that 1 mole of acid reacts with 1 mole of sodium hydroxide.
3. Use the mole ratio to find the number of moles of acid used:
Moles of acid = 0.0039975 mol
4. Calculate the mass of the acid used:
Mass of acid = moles of acid × molar mass of acid
Mass of acid = 0.0039975 mol × 169.7 g/mol = 0.679 g (rounded to three decimal places)
Therefore, the mass of the acid to be used is approximately 0.679 grams.