A 6% (w/v) solution of urea is isotonic with

When a solution of urea(6%) is isotonic with a solution of glucose,
$\begin{array}{l}{\pi }_{glu\cos e}={\pi }_{urea}or{C}_{glu\cos e}=C_{urea}\\ \therefore \left(\frac{w_g\times 1000}{m_g\times 100}\right)=\left(\frac{w_u\times 1000}{m_u\times 100}\right)\end{array}$ 
Where , $w_g$ =mass of glucose=xg; $m_g$ =molecular mass of glucose =180; $w_u$ =mass of urea=6g; $m_u$ =molecular mass of urea=60
$\frac{x\times 1000}{180\times 100}=\frac{6\times 1000}{60\times 100}\Rightarrow x=18g$ 
Thus,18g of glucose are present in 100 mL of solution.
In other words, 1M solution of glucose(18g in 100 mL) is isotonic with 6% solution of urea.