Question

Bài 1: Hãy tính nồng độ mol thu được trong trường hợp sau:

a) 250ml dung dịch có hòa tan 0,1 mol H₂SO₄

b) 400ml dung dịch có hòa tan 20g NaOH

Answer 1

a, \(C_{M_{H_2SO_4}}=\dfrac{0,1}{0,25}=0,4\left(M\right)\)

b, \(n_{NaOH}=\dfrac{20}{40}=0,5\left(mol\right)\)

\(\Rightarrow C_{M_{NaOH}}=\dfrac{0,5}{0,4}=1,25\left(M\right)\)

Answer 2

\(a)C_{M\left(H_2SO_4\right)}=\dfrac{0,1}{0,25}=0,4M\\ b)C_{M\left(NaOH\right)}=\dfrac{\dfrac{20}{40}}{0,4}=1,25M\)

Answer 3

\(a,n_{H_2SO_4}=0,1mol\)

\(V_{H_2SO_4}=0,25l\left(250ml=0,25l\right)\\ C_{M_{H_2SO_4}}=\dfrac{n}{V}=\dfrac{0,1}{0,25}=0,4M\)

`b,` \(n_{NaOH}=\dfrac{m}{M}=\dfrac{20}{40}=0,5mol\)

\(V_{NaOH}=0,4l\left(400ml=0,4l\right)\)

\(C_{M_{NaOH}}=\dfrac{n}{V}=\dfrac{0,5}{0,4}=1,25M\)

Answer 4

a. \(250ml=0,25l\) 

\(C_{MddH_2SO_4}=\dfrac{n_{H_2SO_4}}{V_{ddH_2SO_4}}=\dfrac{0,1}{0,25}=0,4M\) 

\(b.400ml=0,4l\) 

\(n_{NaOH}=\dfrac{m_{NaOH}}{M_{NaOH}}=\dfrac{20}{40}=0,5mol\) 

\(C_{MddNaOH}=\dfrac{n_{NaOH}}{V_{ddNaOH}}=\dfrac{0,5}{0,4}=1,25mol\)