In case of solution,molarity is calculated by using molarity equation M₁V₁ = M₂V₂, we have V₁(before dilution) and V₂ (after dilution) so calculate molarity of the given solution from this equation.
Given that, M₁ = 5 M \(\Rightarrow \) V₁ = 500 mL
V₂ = 1500 mL \(\Rightarrow \) M₂ = M
For dilution, a general formula is
\(\underset { Before \ diution }{ { M }_{ 1 }{ V }_{ 1 }= } \underset { \quad After \ diution }{ { M }_{ 2 }{ V }_{ 2 } } \) 
\(500\times 5M=1500\times M\Rightarrow M=\frac { 5 }{ 3 } =1.66M\)

\(2A+4B\longrightarrow 3C+4D\)
According to the given reaction, 2 moles of A rect with 4 moles of B.
Hence, 5 moles of A will react with 10 moles of
\(b\left( \frac { 5\times 4 }{ 2 } =10moles \right) \)
Limiting reagent decide the amount of product produced. According to the reaction, 4 moles of B produces 3 moles of C.
6 moles of B will produce\(\frac { 3\times 6 }{ 4 } =4.5\) moles of C.

Molality is defined as the number of moles of solute present in 1kg of solvent. It is denoted by m.
Thus Molality (m)
\(=\frac { moles\ of\ solute }{ mass\ of\ solvent } \)
Given that, Mass of solvent (H₂O)
= 500g
= 0.5kg
Weight of HCI
= \(1\times 1+1\times 35.5=36.5g\)
Molar of HCI(solute) 
\(=\frac { 18.25 }{ 36.5 } =0.5\)
\(m=\frac { 0.5 }{ 0.5 } =1m\)

In the given question 0.9g L^-1 means that 1000 mL solution contains 0.9 g of glugose
Number of moles = 0.9g glucose = \(\frac { 0.9 }{ 180 } \)mol glugose
= \(5\times { 10 }^{ -3 }\)mol glugose.
(where molecular mass of glugose(C₂H₁₂O₆) = \(12\times 6+12\times 1+6\times 16=180u)\)
i.e IL solution contains 0.05 mole glugose or the molarity of glucose is 0.005M.