Please help and show how you find the answer so I can see how you got it. I know

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Please help and show how you find the answer so I can see how you got it. I know how to do question 1 and 4. I am stuck on the rest. Thanks!

1. A saline sodium citrate buffer is a commonly used reagent in biochemistry/molecular biology. In this reagent the "saline" refers to aqueous sodium chloride present at a molality of 3.000 M. If the density of a 3.000 M aqueous solution of Naci is 1.115 g/mL, what is the molality of this solution? (sHow YOUR woRK!) d) 3.19 c) 3.12 e) 3.31 a) 3.00 b) 3.04 Ricin is a lethal substance extracted from the seeds of the castor oil plant and was used in assassination attempts during the Cold War, as well as being "popularized" by of TV shows over the years. The lethal dose in humans is around 22 ug per 1 kilogram body weight. Ricin is a protein with a molar mass of about 65 000 g/mol. If an average man weighing 85 kg is injected with ricin that is initially all constrained within the 5.0 L of blood in his body, what is the approximate molarity of ricin in his blood required to reach the lethal dose? (Show your work!) a) 3.2 x 10 M b) 5.8 x 10 M c) 7.1 x 10- M d) 8.3 x 10 M e) 4.7 x 10 M 10.0 g of sodium metal and 20.0 g of gas react together to produce a white crystalline substance that is subsequently dissolved to a total volume of 250 mL. molar in water What is the of that resulting solution? a)88 b)1.39 c)1.74 d)2.51 e)3.48 4. On the night of the sinking of the Titanic, the ocean water was around 1°c, roughly the temperature at which oceanic water freezes. Given that the freezing point depression is 2°C, approximately how many grams of sodium chloride are there in 1 kg of ocean water, considering the ocean consists of primarily sodium ions, chloride ions and water? a)15.8 b)18.4 c) 23.9 d) 27.1 e) 31.5 5. A 1.0 m aqueous solution of potassium sulfate has a freezing point of -4.3°C. What is the numerical difference between the actual van't Hoff factor and the ideal van't Hoff factor? a)0.3 b)0.7 c 1.0 d)2.3 e) 3.0

Explanation / Answer

Q1.

M = 3 M of saline, NaCl,

D = 1.115 g/mL, find molality

First, define molality

molality (m) = mol of solute / kg of solvent

solute = NaCl

solvent = H2O

now, assume a basis, 1 Liter of solution, so

M = mol/V

3 M = mol/ 1 L

mol = (3)(1) = 3 mol of NaCl per liter of solution

mass of NaCl = mol*MW = 3*58.44 = 175.32 g of NaCl

then.. calculate mass of solution

mass of solution = mass of water + mass of salt

D = mass/Vol

D = 1.115 g/mL

Vol = 1 liter or 1000 mL

Mass = D*V = 1.115*1000 = 1115 g

Now, calculate mass of water

mass of water = mass of solution - mass of salt = 1115-175.32 = 939.68 g of water

change to kg, 939.68 g = 0.93968 kg

finally, calcualte molality

molality = mol / kg = 3/0.93968 = 3.19257 molal

fro the list, best option is D

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