Home
Class 11
CHEMISTRY
An exhausted zeolite bed was revived by ...

An exhausted zeolite bed was revived by 250 L of NaCl solution containing `50gL^(-1)` of NaCl solution. How many litres of hard water of hardness 250 ppm can be softened on the zeolite bed?

Text Solution

AI Generated Solution

To solve the problem step by step, we will follow the outlined procedure to determine how many liters of hard water with a hardness of 250 ppm can be softened using the zeolite bed that has been revived with a NaCl solution. ### Step 1: Understand the Problem We need to find out how many liters of hard water can be softened by a zeolite bed that has been regenerated with a specific volume of NaCl solution. ### Step 2: Gather Given Data - Volume of NaCl solution (V2) = 250 L - Concentration of NaCl solution = 50 g/L ...
Promotional Banner

Topper's Solved these Questions

  • STOICHIOMETRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 3.7|5 Videos

  • STOICHIOMETRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises Subjective|31 Videos

  • STOICHIOMETRY

    CENGAGE CHEMISTRY ENGLISH|Exercise Ex 3.5|6 Videos

  • STATES OF MATTER

    CENGAGE CHEMISTRY ENGLISH|Exercise Exercises (Ture False)|25 Videos

  • THERMODYNAMICS

    CENGAGE CHEMISTRY ENGLISH|Exercise Archives (Subjective)|23 Videos

Similar Questions

Explore conceptually related problems

The hardness of 10^5 L of a sample of H_2O was completely removed by passing through a zeolite softener. The bed on exhaustion required 500 L of NaCl solution containing 15gL^(-1) of NaCl for regeneration. Calculate the hardness of the sample of water.

How many grams of NaOH should be added to water to prepare 250 ml solution of 2 M NaOH?

Semimolar solution contains how many moles of solute in 1L of solution?

How many moles of sodium chloride present in 250 mL of a 0.50 M NaCl solution ?

If 250 ml of 0.25 M NaCl solution is diluted with water to a volume of 500 ml, the new concentration of solution is:

1.325 g of anhydrous sodium carbonate are dissolved in water and the solution made up to 250 mL. On titration, 25 mL of this solution neutralise 20 mL of a solution of sulphuric acid. How much water should be added to 450 mL of this acid solution to make it exactly N/12 ?

12. g of an impure sample of arsenious oxide was dissolved in water containing 7.5 g of sodium bicarbonate and the resulting solution was diluted to 250 mL . 25 mL of this solution was completely oxidised by 22.4 mL of a solution of iodine. 25 mL of this iodine solution reacted with same volume of a solution containing 24.8 g of sodium thiosulphate (Na_(2)S_(2)O_(3).5H_(2)O) in one litre. Calculate teh percentage of arsenious oxide in the sample ( Atomic mass of As=74 )

A sample of hard water is found to contain 40 mg of Ca^(2+) ions per litre. The amount of washing shoda (Na_(2)CO_(3)) required to soften 5*0 L of the sample would be

0.0093 g of Na_(2)H_(2)EDTA.2H_(2)O is dissolved in 250 mL of aqueous solution. A sample of hard water containing Ca^(2+) and Mg^(2+) ions is titrated with the above EDTA solution using a buffer of NH_(4)OH+NH_(4)Cl using eriochrome balck- T as indicator. 10 mL of the above EDTA solution requires 10 mL of hard water at equivalent point. another sample of hard water is titrated with 10 mL of above EDTA solution using KOH solution (pH=12) . using murexide indicator, it requires 40 mL of hard water at equivalence point. a. Calculate the ammount of Ca^(2+) and Mg^(2+) present in 1L of hard water. b. Calculate the hardness due to Ca^(2+) , mG^(2+) ions and the total hardness of water in p p m of CaCO_(3) . (Given MW(EDTA sal t)=372 g mol^(-1),MW(CaCO_(3))=100 gmol^(-1) )

If 250 cm^3 of an aqueous solution containing 0.73 g of a protein A of isotonic with one litre of another aqueous solution containing 1.65 g of a protein B, at 298 K, the ratio of the molecular masses of A and B is ________ xx 10^(-2) ( to the nearest integer).