What is the mole concentration of 58.44g of NaCl in 1 L of water?
Molarity = moles of solute/Liters of solution
Get moles NaCl.
58.44 grams NaCl (1 mole NaCl/58.44 grams)
= 1 mole NaCl
Molarity = 1 mole NaCl/1 liter
= 1 M NaCl
What is the concentration M of a NaCl solution prepared by dissolving 3.8 g of NaCl in sufficient water to give 245 mL of solution?
What has a greater boiling point elevation when dissolved at a concentration of 1mol in water with NaCl or MgCl2?
In the same volume of water, one mole of MgCl2 will give rise to a greater boiling point elevation. This is explained by the fact that boiling point elevation is a colligative property, that is, the relative amounts of the constituents are important and not their identity. We can determine by inspection that, upon dissociation, more ions will be produced by MgCl2 than NaCl since there are more atoms in the MgCl2 molecule. So, for…
Mole = Mass/RMM Ar for Na=23 & Ar for Cl = 35.5 Therefore 1mole = Mass / (35.5+23) 1 mole of NaCl is 58.5g Mole = Mass/RMM Ar for Na=23 & Ar for Cl = 35.5 Therefore 1mole = Mass / (35.5+23) 1 mole of NaCl is 58.5g But that can be misleading. NaCl does not form into molecules (and salts in general do not). If you dissolve 58.5g of NaCl into water, it will…
Mole percent is the mole fraction (X) for a component of a mixture times 100. If you have a solution of 5.0g NaCl and 25.0g H2O, you can find the mole fraction and mole percent for each. The sum of all percents of all of the components of a mixture equals 100. Determine the moles (n) of NaCl and H2O. 5.0g NaCl x 1mol NaCl/58.44g NaCl = 0.086mol NaCl 25.0g H2O x 1mol H2O/18.016g H2O…
What volume of a 2.10 M NaCl solution is needed for a reaction that requires 200 g of NaCl answer in units of L?
An impure sample of table salt that weighed 0.8421 g when dissolved in water and treated with excess AgNO3 formed 2.044 g of Ag-Cl what is the percentage of Na-Cl in the impure sample?
Saline solution is 9 grams of sodium chloride (NaCl) dissolved in 1000ml water. The mass of 1 millilitre of normal saline is 1.0046 gram at 22 °C. The molecular weight of sodium chloride is approximately 58.5 grams per mole, so 58.5 grams of sodium chloride equals 1 mole. Since normal saline contains 9 grams of NaCl, the concentration is 9 grams per liter divided by 58.5 grams per mole, or 0.154 mole per liter. Since…
The formula weight of NaCl is 58.44 which means 58.44 g = 1 mole. Since 1000 mg = 1 g, this gives 58.44 g/mole x 1000 mg/g = 5.844 E+4 mg/mole. Then, 52.1 mg NaCl x 1 mole NaCl/5.844E+4 mg = 8.92 E-4 mole. See, this is just done by using the definitions of the units and the factor label method to analyze the conversion.
If by salt, you mean NaCl, then 50 g NaCl x 1 mole NaCl/58.4 g = 0.856 moles 0.856 moles NaCl/0.2 kg= 4.28 molal = 4.28 m = 4.3 m (2 sig figs) NOTE: You cannot calculate the molarity (M) without knowing the final volume. Adding 50 g salt to 200 ml will NOT end in a final volume of 200 ml. It will be somewhat greater than that.
sugar is a covalent compound where as salt is an ionic compound,so salt while dissolving in the water splits. This is mainly dependent of the 'total dissolved partical' concentration (mol/L): Sugar (C6H12O6): 1 mole particles per 180 g (for 1 mole sugar) Salt (NaCl) 2 mole particles per 58.5 g (for 1 mole salt) = 6 mole particles per about 180 In water the freezing point will be lowered by 1.86oC per mole particles dissolved…