The specific gravity of seawater is 1.025, not the usual 1.00 for water, and is

Question

The specific gravity of seawater is 1.025, not the usual 1.00 for water, and is 3.5% salt (mostly NaCl) by mass. This increased density is due primarily to the dissolved NaCl in the seawater. Give the density (kg/m3) of the seawater, and then the molarity, molality, ppm (per mass), mole fraction, and TDS of the NaCl in the water, as if it were the only dissolved component. We did not discuss TDS in class, because I want you to find its meaning through a quick search.  answers: r = 1025 kg/m3, conc = 0.614 M, 0.621 m, 35,000 ppm, x = 0.0110, TDS 35,900 mg/L.

Explanation / Answer

specific gravity of sea water containing 3.5% salt = 1.025

density= specific gravity* desnity of water= 1.025*1000 kg/m3= 1025 kg/m3

basis : 100 gm of seal water. It contains 3.5 % NaCl. Hence mass of NaCl= 3.5 gm, moles of NaCl= mass/molar mass= 3.5/58.5=0.0598 mole, volume of 100gm of sea water= mass/density in g/cc= 100/1.025= 97.56 ml, since 1000ml= 1L, 97.56ml= 97.56/1000L= 0.09756L, molarity of NaCl= moles of solute/liter of solution = 0.0598/0.09756=0.613M.

mass of water in 100gm= mass of solution- mass of NaCl =100-3.5= 96.5 gm, 1000gm= 1kg, 96.5 gm= 96.5/1000 kg =0.0965 kg, molality= moles of solute/ kg of solvent =0.0598/0.0965=0.612m

3.5 gm of NaCl is present in 100gm or 3.5 gm/100gm= 3.5*104/106 gm/gm =35000 gm/106 gm = 35000ppm

mole of water = mass of water/ molar mass= 96.5/18, moles of NaCl =0.0598

mole fractino of NaCl= moles of NaCl/total moles = 0.0598/(0.0598+96.5/18) =0.011

TDS= 3.5/97.56 gm/ml=0.0359 gm/ml= 0.0359*1000gm/L = 35.9 gm/L= 35.9*1000mg/L= 35900mg/L

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