
A 5% solution of cane sugar (molecular weight 342g) is isotonic with 0.877% solution of urea. Find the molecular weight of urea.
Answer
485.1k+ views
Hint: Isotonic solutions have the same concentration. Find the molarity of cane sugar solution and equate it to that of urea solution.
Complete answer:
We have been given a 5% weight by volume solution of cane sugar. We know this is weight by volume because the molecular weight of cane sugar is provided to us. Now what does it actually mean?
Well, it means we have 5g of cane sugar in 100ml of water. To know its actual concentration, we have to find its molarity. Molarity is defined as the number of moles of solute in one litre of solution. By unitary method we can calculate it as,
\[\begin{align}
& 100\operatorname{ml}\to 5\operatorname{g} \\
& 1000\operatorname{ml}\to 50\operatorname{g} \\
\end{align}\]
Now to convert 50g of cane sugar into moles we have to divide it by its molecular weight which is already given as 342g.
\[\text{No}\text{. of moles of cane sugar}\to \dfrac{50g}{342g}=0.1461\]
Now we have found the molarity of the given cane sugar solution, which is 0.1461M.
As the urea solution is isotonic to cane sugar solution, it means, they have the same molarity. It is given that urea solution is a 0.877% weight by volume solution, which means 0.877g of urea is present in 100ml of solution, that implies 8.7g of urea is present in 1000ml of the solution; which we calculated by unitary method. It is shown as follows:
\[\begin{align}
& 100\operatorname{ml}\to 0.877\operatorname{g} \\
& 1000\operatorname{ml}\to 8.77\operatorname{g} \\
\end{align}\]
As we know, the number of moles present in 1000ml of urea solution is the same as that of the cane sugar solution and that is equal to 0.1461. We can therefore calculate the molecular weight of urea from the following formula:
\[\begin{align}
& \text{Molecular weight}=\dfrac{Given\text{ mass}}{No\text{. of moles}}=\dfrac{8.7}{0.1461} \\
& \text{Molecular weight}=59.548\operatorname{g}\approx 60\operatorname{g} \\
\end{align}\]
Hence, the molecular weight of urea is 60g.
Note:
The information about a particular solution need not always be given in weight by volume style. It can also be in volume by volume or number of moles by volume or anything else. A student has to look out for these hints given in the question itself.
Understanding the basic concepts about the concentration of a solution such as molarity, molality and mole fraction are crucial in order to solve these questions.
Complete answer:
We have been given a 5% weight by volume solution of cane sugar. We know this is weight by volume because the molecular weight of cane sugar is provided to us. Now what does it actually mean?
Well, it means we have 5g of cane sugar in 100ml of water. To know its actual concentration, we have to find its molarity. Molarity is defined as the number of moles of solute in one litre of solution. By unitary method we can calculate it as,
\[\begin{align}
& 100\operatorname{ml}\to 5\operatorname{g} \\
& 1000\operatorname{ml}\to 50\operatorname{g} \\
\end{align}\]
Now to convert 50g of cane sugar into moles we have to divide it by its molecular weight which is already given as 342g.
\[\text{No}\text{. of moles of cane sugar}\to \dfrac{50g}{342g}=0.1461\]
Now we have found the molarity of the given cane sugar solution, which is 0.1461M.
As the urea solution is isotonic to cane sugar solution, it means, they have the same molarity. It is given that urea solution is a 0.877% weight by volume solution, which means 0.877g of urea is present in 100ml of solution, that implies 8.7g of urea is present in 1000ml of the solution; which we calculated by unitary method. It is shown as follows:
\[\begin{align}
& 100\operatorname{ml}\to 0.877\operatorname{g} \\
& 1000\operatorname{ml}\to 8.77\operatorname{g} \\
\end{align}\]
As we know, the number of moles present in 1000ml of urea solution is the same as that of the cane sugar solution and that is equal to 0.1461. We can therefore calculate the molecular weight of urea from the following formula:
\[\begin{align}
& \text{Molecular weight}=\dfrac{Given\text{ mass}}{No\text{. of moles}}=\dfrac{8.7}{0.1461} \\
& \text{Molecular weight}=59.548\operatorname{g}\approx 60\operatorname{g} \\
\end{align}\]
Hence, the molecular weight of urea is 60g.
Note:
The information about a particular solution need not always be given in weight by volume style. It can also be in volume by volume or number of moles by volume or anything else. A student has to look out for these hints given in the question itself.
Understanding the basic concepts about the concentration of a solution such as molarity, molality and mole fraction are crucial in order to solve these questions.
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