sodium nitrate + water $\rightarrow$ saturated solution
At $40^{\circ} C$ 104 g + 100 g $\rightarrow$ 204 g At $40^{\circ} C$
204 g of saturated solution contains 104 g of sodium nitrate
25.5 g of saturated solution contains =?
$\begin{array}{l}
=\frac{25.5g \times 104g}{204g} \\
=13.00 g
\end{array}$
Answer (1)
Set up a proportion to relate the amount of sodium nitrate to the amount of saturated solution.
Express the proportion as: 25.5 x = 204 104 , where x is the amount of sodium nitrate in 25.5 g of saturated solution.
Solve for x by multiplying both sides by 25.5: x = 204 104 × 25.5 .
Calculate the value of x : x = 13 . Therefore, the final answer is 13 .
Explanation
Understanding the Problem We are given that at 4 0 ∘ C , 104 g of sodium nitrate dissolves in 100 g of water to form 204 g of saturated solution. This means that 204 g of saturated solution contains 104 g of sodium nitrate. We want to find out how much sodium nitrate is present in 25.5 g of saturated solution.
Setting up a Proportion Let x be the amount of sodium nitrate in 25.5 g of saturated solution. We can set up a proportion to solve for x . The proportion is: grams of saturated solution grams of sodium nitrate = 204 104
Writing the Proportion Now we can write the proportion with the unknown x : 25.5 x = 204 104
Solving for x To solve for x , we multiply both sides of the equation by 25.5: x = 204 104 × 25.5 Now, we perform the calculation: x = 204 104 × 25.5 = 204 2652 = 13 So, there are 13 grams of sodium nitrate in 25.5 g of saturated solution.
Final Answer Therefore, 25.5 g of saturated solution contains 13 g of sodium nitrate.
Examples
Understanding the concentration of solutions is crucial in many real-world applications. For example, in medicine, it's important to know the exact concentration of a drug in a solution to administer the correct dosage. In cooking, the concentration of salt in a brine solution affects the flavor and preservation of food. In chemistry, knowing the concentration of reactants is essential for controlling chemical reactions. By calculating the amount of solute in a given amount of solution, we can ensure accuracy and consistency in these and many other applications.
