Determne the boiling point of 16.3g of calcium chloride dissolved in 250. g water.
My work:
16.3 g CaCl2 *(1 mole CaCl2/110.984 g Cacl2)= 0.147 mol CaCl2
M= 0.147/0.250kg =0.588 mCaCl2 ... From here, How do you find out how many ions CaCl2 has. Calcium has a 2+ charge and Cl is - charge. I don't get how together it is 3 ions. I need this piece of info because I need to multiply 0.588 by 3 and I'm not sure how my teacher got 3, and I'm confused. It should = 1.76 M CaCl2.
To find boiling point... I know how to do the rest. I'm just confused on the IONS part. How to determine that?
#2) Another question is similar. Calculate the freezing pt of sol'n that contains 32.7 g of sodium acetate dissolved in 250. g of water. I know how to find the freezing point but the number I'm confused of to figure out my answer is the #of ions. It's supposed to be 2 ions (it's the particle molality)... ... I know Sodium has a + charge and Acetate is a - charge. How is it 2 ions though if it cancels out???
a. CaCl2=3ions; 1 Ca+2, two Ca- ions. Now to find bp, shouldn't you be using molality, not Molarity?
b. Na+ Acetate- ion, two ions.
ions do not cancel out, you need a new brand of wine. Ions are particles, you count them, you dont cancel them out. If you had 100 males, and 100 female, that is 200 persons, they did not cancel out and become couples. Sodium acetate is a couple, but when it dissolves, it is two ions.
To determine the number of ions in a compound, you need to consider the formula of the compound and its charges. In the case of calcium chloride (CaCl2), you correctly identified that calcium has a 2+ charge and chloride has a 1- charge.
Since there are two chloride ions (Cl-) for every one calcium ion (Ca2+), the total number of ions contributed by calcium chloride is 3 (1 calcium ion and 2 chloride ions). Therefore, when calculating the molality (m) of calcium chloride, you multiply the molality of the compound (0.588 mCaCl2) by 3 to account for the three ions. This gives you a final result of 1.764 mCaCl2.
Similarly, for sodium acetate (NaC2H3O2), sodium has a 1+ charge and acetate has a 1- charge. In this case, there is one sodium ion (Na+) and one acetate ion (C2H3O2-) for each formula unit of sodium acetate. Therefore, the total number of ions contributed by sodium acetate is 2 (1 sodium ion and 1 acetate ion). When you calculate the molality of sodium acetate, you multiply the molality of the compound by 2 to account for the two ions.
It's important to note that in ionic compounds, the charges of the ions have to balance out overall to form a neutral compound. In the case of calcium chloride, calcium's 2+ charge balances out the 2- charge from the two chloride ions, resulting in a neutral compound. Similarly, sodium's 1+ charge balances out the 1- charge from the acetate ion in sodium acetate.