The given question is incomplete. The complete question is:
When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.
Answer: The vant hoff factor for sodium chloride in X is 1.9
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
= freezing point constant
i = vant hoff factor = 1 ( for non electrolyte)
m= molality =
Now Depression in freezing point for sodium chloride is given by:
= Depression in freezing point
= freezing point constant
m= molality =
Thus vant hoff factor for sodium chloride in X is 1.9
Answer:
Different types of isotopes are used for different materials or objects. For radiometric dating, uranium-235 is considered best for it while carbon-14 is used for dating of rocks. It is also used for dating of wood samples.
Explanation:
Carbon-14 and uranium-235 are used for different materials or objects for measuring the age of these materials. These two isotopes are radioactive in nature which means they emit gamma radiations which allow us to find the age of different objects. Carbon-14 has a low half life so it can be used for those objects which are present before thousands of years while uranium-235 is used for materials which are millions of years old due to high half life.
Answer : The energy required to heat of 1.50 kg iron is,
Explanation :
Formula used :
or,
where,
Q = heat required = ?
m = mass of iron = 1.50 kg = 1500 g
c = specific heat of iron =
= initial temperature =
= final temperature =
Now put all the given value in the above formula, we get:
Therefore, the energy required to heat of 1.50 kg iron is,
Aluminum would be the answer
Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is
To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.