What molarity of nitric acid (HNO3) was used if 2.00 L must be used to prepare 4.5 L of a 0.25 M HNO3 solution?

The molarity of (HNO₃) that was used if 2.00 L must be used to prepare 4.5 L of a 0.25M HNO₃ solution is 0.563 M

calculation

This is calculated usind M₁V₁=M₂V₂ formula

where,

M₁ (molarity ₁) = ?

V₁ (volume ₁) = 2.00 L

M₁ (molarity ₂) = 0.25M

V₂ (volume₂) = 4.5 L

make M₁ the subject of the formula by diving both side of the formula by V₁

M₁ is therefore = M₂V₂/V₁

M₁ = [ (0.25 M x 4.5 L) / 2.00 L ] = 0.563 M