Molality of acid NOT Molarity?

Question

2.29 mL of concentrated sulfuric acid (specific gravity 1.84) is added to mass of 0.1978 kg of water. The sulfuric acid solution is then titrated with NaOH and no. of mol of NaOH that reacted with H2SO4 is 0.0092 mol.

How do you calculate the molality of the acid produced by the dilution?

Can anyone pls help? Thanks a lot

I thought 2 mol of NaOH react with 1 mole of H2SO4?

Answer 1

Molality = moles of solute/ kg of solvent

specific gravity of H2SO4 = density of H2SO4

mass of sulfuric acid = specific gravity*volume = 1.84*2.29
= 4.2136 mg

mols of sulfuric acid = mass/molecular weight = 4.2136/98 = 0.04299 mol

since 2 mol NaOH reacts with 1 mol H2SO4
0.0092 mol mol NaOH reacts with 0.0046 mol mol H2SO4

thus unreacted H2SO4 = 0.04299 -0.0046 mol = 0.03839mol

weight of solvent water = 0.1978 kg

molality of the resulting solution = 0.03839/0.1978 = 0.194

Answer 2

solution provided by Som is correct with minor correction that 2 moles of NaOH reacts with 1 mol of sulfuric acid

Answer 3

molality of a substance is no.of moles / molecular mass.
symbol : small "m"
therez a formula similar to molarity :
m1v1 = m2v2
m=no of moles/molar mass
substistute that back in the formula i juss gave..but the values given in the ques. find the one wich is asked

Answer 4

the 1st element you desire to do is be sure out what selection moles of HCl you have interior the respond. it fairly is not an comparable using fact the mass fraction. To get mole fraction, you besides mght decide to envision out what selection moles of water you have. to try this you desire to divide the mass of the substance by way of its molecular weight. some ingredient is straight away forward.

Answer 5

molality=moles/kg

molarity= moles/ltr